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Answer 1 · 2017-04-05T14:38:36

$27.4g (3s.f.)$

$2C_4H_10+13O_2->10H_2O+8CO_2$

relative formula mass $(M_r)$ of $2C_4H_10$ :

$A_m(C)=12$
$A_m(H)=1$

$M_r(2C_4H_10) = 2*(4*12+10*1)$
$=2*58$
$=116$

$M_r(2C_4H_10) = 116$

$M_r(8CO_2) = 8*(12+(2*16)) = 8*44$
$=352$

$M_r(2C_4H_10) = 116$
$M_r(CO_2) = 44$

$(M_r(2C_4H_10))/(M_r(CO_2)) =$ mass of butane needed/ $(83.0g)$

$116/352=29/88=$ mass of butane needed/ $(83.0g)$

mass of butane needed = $29/88 * 83.0g$
$= 27.4g (3 s.f)$